- 10 Star-delta transformation 59
(^1) .,, I 1.67s S 5~
- i ] ~] I
2.5~
10V 10~~-]
7.5D.
3 A 1.67~
I --!!L5gZ 10VB
Figure 3.34 Figure 3.35IL10~The equivalent resistance between A and B is then given by
RAB = R a + 10 • 10/(10 + 10) = 1.67 + 5 = 6.67 fZ
The current IL in Fig. 3.32 is the same as that in Fig. 3.35 and is obtained by
current division. Since the two parallel connected resistors are equal in value,
the total current I will divide equally between them. Now
I = V/RA, = (10/6.67)A - 1.5 A
so that IL = 0.75 A.3.10 STAR-DELTA TRANSFORMATION
We can use the same diagrams of Figs 3.29 and 3.30 and obtain the reverse
transformation by considering Equations (3.21), (3.22) and (3.23) above.
Multiplying Equation (3.21) by Equation (3.22), Equation (3.22) by Equation
(3.23) and Equation (3.23) by Equation (3.21) in turn, we obtain
R1R2 = RazZRz3R3]/[R12 + R23 -+- R31] 2 (3.24)
R2R3 = R232R31R,2//IR12 + R23 + R31] 2 (3.25)
R3R1 = R312R~2R23//[R12 + R23 + R3~] 2 (3.26)
Adding Equations (3.24), (3.25) and (3.26) gives
R~R2 + R2R3 + R3R1 = R12R23R31[R~2 + R23 + R3~]/[R~2 + R23 + R31] 2
= R12R23R31/(R12 + R23 + R31)
Now R23R3]/(R12 + R23 + R31) - R3 from Equation (3.23) so
RIR2 + RzR3 + R3R1 = RlzR3Dividing both sides by R3 we see that
R12 = R1 + R2 + R1Rz/R3 (3.27)
Similarly, by noting that
R31Rlz/(R12 + R23 + R31) = R1